BITSAT 2021 | Work, Energy and Power Question 6 | Physics

A force F = a + bx acts on a particle in the x-direction where a and b are constants. The work done by this force during a displacement from x = 0 to x = d is

$$\left( {a + {{bd} \over 2}} \right)d$$

$$\left( {ad + {{bd} \over 2}} \right)$$

$$\left( {{{ad + bd} \over 2}} \right)$$

$$(ad + b){d \over 2}$$

BITSAT 2021

MCQ (Single Correct Answer)

-1

The bob of a pendulum of length 2 m lies at P, when it reaches Q, it losses 10% of its total energy due to air resistance.

BITSAT 2021 Physics - Work, Energy and Power Question 9 English

The velocity of bob at Q is

6 m/s

1 m/s

2 m/s

8 m/s

BITSAT 2020

MCQ (Single Correct Answer)

-1

A ball of mass 0.5 kg is thrown up with initial speed 16 ms^$$-$$1 and reaches maximum height of 9 m. How much energy is dissipated by air drag acting on the ball during the ascent?

199 J

19.9 J

20.9 J

9.9 J

BITSAT 2020

MCQ (Single Correct Answer)

-1

A time dependent force F = (8t) acts on a particle of mass 2 kg. If the particle starts from rest, the work done by the force during the first 1s will be

0.4 J

4 J

19 J

4.5 J

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